# Statically Indeterminate System

If we draw the free body diagram of bar ACB of Fig. 3.1, we get as per Fig. 3.2. The weight of the bar is neglected. So, P, 9, x, and y are known.

Here Number of unknowns = (RAX, RAy and RBy) = 3

Number of equations available =  (Σ fx = 0, Σ fy = 0 , Σ M = 0) = 3

Number of unknowns = Number of equations available.

So the problem in Fig. 3.1 is called statically determinate problem.

Now let us draw the free body diagram of the bar ACB of Fig. 3.3. The free body

diagram is shown in Fig. 3.4.

As before weight of the bar ACB is neglected. P, θ, x and y are known quantities. Here

Number of Unknowns = (RAx, RAy‘ RBx, RBy) = 4 Number of equations available from statics

= 3 (Σ fx = 0, Σ fy = 0 , Σ M = 0)

Number of unknowns > Number of equations available.

So, the problem in Fig. 3.3 is statically indeterminate.

Degree of Indeterminacy = Number of unknowns – Number of equations available

= 4 – 3 = 1 (For Fig. 3.3).

Thus, Fig. 3.3 is called statically indeterminate problem to the degree one.

The free body diagram of the bar ACB of Fig. 3.5 is shown in Fig. 3.6.

Here, Number of unknowns = (RAx, RAy‘ RBx, RBy and MB) = 5

Number of equations available from statics = 3

. . Degree of indeterminacy = 5-3 = 2

Solution of Statically Indeterminate Problem

For the solution of a statically indeterminate problem of degree of indeterminacy m, ‘m’ number of supplimentary independent equations are to be set up from the consideration of the deformation of the structure. The following example describes the solutions (Fig. 3.7).

Here,

Number of unknowns = 4

Number of equations available = 3

Degree of indeterminacy = 4 – 3 = 1.

If we neglect the weight of the bar the vertical reactions RAy and Ray addition will be

zero.

From Fig. 3.9,

Σ fx = 0  =>

RAX + P – RBX = 0

RAX – RBX = – P

The problem is now to find another equation.

For this we consider the elastic deformation of the bar ACB. Assuming that the two supports are·perfectly rigid, the total deformation for the total length (a + b) is zero.

δ1 = -RAx x a / AE

δ2 = – RBx x b / AE

δ1 + δ2 = 0

RAx x a + RBx x b = 0

RBx = – RAx x a /b

Solving for (3.1) and (3.2), we get

(3.1) =>             RAx x a / b = p

RAx ( a +b / b) = p

RAx = pb / a+b

(3.2) =>            RBx x a /b = – pa / a +b

The direction of R ax will be opposite to that of P.